-1
$\begingroup$

In my book, the graph for isothermal expansion for 2 steps and n steps is shown as this

enter image description here

I can't understand the graph for both compression and expansion. Why is external pressure changing in going from $V_1$ to $V_2$. I thought the external remained constant in irreversible reaction and why is external pressure different for both compresdion and expansion?

$\endgroup$

closed as off-topic by Mithoron, A.K., Mathew Mahindaratne, a-cyclohexane-molecule, Tyberius Aug 13 '18 at 2:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

They are trying to show what happens if you expand or compress a gas in one or more steps at constant externally applied pressure (irreversible process), particularly when the number of steps becomes very large. When you expand the gas in this framework, you suddenly drop the external pressure, and then hold it constant while you let the gas equilibrate at the new lower pressure. When you compress the gas in this framework, you suddenly raise the external pressure, and then hold it constant while you let the gas equilibrate at the new higher pressure. So, of course, the expansion steps are going to look different from the compression steps on the graph. The idea is that, as you increase the number of steps (between two fixed end points), the cumulative amount of irreversibility becomes less, as you approach a reversible process.

$\endgroup$
0
$\begingroup$

Yes, you are right that external pressure remains constant in an irreversible process. But in this case expansion and compression is done in two step. In case of two-step expansion or compression, the process is carried out by passing the system through an intermediate state.

Let us first consider two-step expansion of $1\ \pu{mol}$ of an ideal gas from ($\mathrm{P_1,\ V_1,\ T_1}$) to ($\mathrm{P_2,\ V_2,\ T_2}$). Therefore to carry out the change in a two-step, at first step we put an external pressure $\mathrm{P_i}$ over the piston which is less than $P_1$ but greater than zero. Now on the removal of the stops fitted upon the piston, the system experiences a sudden change and attains the state ($\mathrm{P_i,\ V_i,\ T_i}$). The external pressure remains constant in this step.

Now in the second step, the external pressure is made equal to $\mathrm{P_2}$ which is lower than $\mathrm{P_2}$ but greater than zero. Now here also on the removal of the stops, the system experiences sudden change and the gas molecules push the mass kept over the piston upwards. As a result, the system experiences a sudden change and the internal pressure becomes equal to $\mathrm{P_2}$ i.e the system attains the state ($\mathrm{P_2,\ V_2,\ T_2}$) i.e. attains the final state. The external pressure remains constant at $\mathrm{P_2}$ throughout the second step.

Therefore in a two-step expansion, the external pressure in a particular step remains constant throughout that process.

But in the overall process, it doesn't remain constant. If the external pressure at 2nd step remains the same as that of in 1st step, then the overall process comes to a stop.

Because expansion or compression takes place due to the urge of the gas molecules to attain the pressure equilibria between internal pressure and external pressure. Now in two-step expansion at 1st step the external pressure is Pi and internal pressure is $\mathrm{P_1}$. Due to this reason on the removal of the stops the gas molecules, due to the presence of the urge to attain the pressure equilibria push up the piston upwards to a certain height until the internal pressure becomes $\mathrm{P_i}$. After attaining $\mathrm{P_i}$ pressure, the system attains the pressure equilibrium i.e both internal and external pressure is $\mathrm{P_i}$.

Therefore in this stage, due to the pressure equilibria, the position of the piston remains same even on removal of the stops. So the process also stops. Now to proceed the process furthermore to achieve our desired final state, we have to disturb again the equilibrium between the external pressure and internal pressure so that the system regain its urge to attain the pressure equilibrium. Due to this reason in two-step expansion, the external pressure is changed after completion of one step.

In a similar manner, the variation of external pressure in two-step compression can also be explained.

$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE! Please note the edits made especially the use of \$\mathrm{}\$ for equations and variables and \$\pu{}\$ for units. We also use \$ce{}\$ for presenting chemical formulas and equations. $\endgroup$ – A.K. Aug 12 '18 at 13:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.